Physics 201: Lecture 23, Pg 1 Lecture 23 l Exam 3: 2103 Chamberlin Hall, B102 Van Vleck & quiet room...
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Transcript of Physics 201: Lecture 23, Pg 1 Lecture 23 l Exam 3: 2103 Chamberlin Hall, B102 Van Vleck & quiet room...
Physics 201: Lecture 23, Pg 1
Lecture 23
Exam 3: 2103 Chamberlin Hall, B102 Van Vleck & quiet room CH Sections 604 605 606 609 610 611 (TA: Moriah T., Abdallah
C., Tamara S.) VV Sections 602 603 607 608 612 (TA: Eric P., Dan C., Zhe D.) Format: Closed book, three 8 x11” sheets, hand written Electronics: Any calculator is okay but no web/cell access Quiet room: Test anxiety, special accommodations, etc.
Chapters Covered Chapter 9: Linear momentum and collision (not 9.9) Chapter 10: Rotation about fixed axis and rolling Chapter 11: Angular Momentum (not 11.5) Chapter 12: Static equilibrium and elasticity
Physics 201: Lecture 23, Pg 2
Basic Concepts and Quantities
Momentum, Angular Momentum Linear Momentum, Impulse Angular Momentum (magnitude and direction)
Torque Collisions: Elastic & Inelastic Center of Mass Rotational Motion (1-axis) Angular displacement (Δ) / Velocity(), Acceleration (). Moments of Inertia Rotational Kinetic Energy Conservation Laws: Energy, Momentum, Angular Momentum Static Equilibrium Elasticity: Young’s, Shear, Bulk Modulus
Physics 201: Lecture 23, Pg 3
Chapter 9 : Momentum and Momentum Conservation
Ix =
Ix
Physics 201: Lecture 23, Pg 4
Chapter 9
Physics 201: Lecture 23, Pg 5
Center of Mass
Chapter 9
M
rmrmrm
m
rmr N
ii
N
iii
CM
332211
1
1
M
vmvmvm
m
vmv N
ii
N
iii
CM
332211
1
1
Physics 201: Lecture 23, Pg 6
Chapter 10
At a point a distance R away from the axis of rotation, the tangential motion:
s (arc) = R vT (tangential) = R
aT = R
R
v = R
s
)(2
21
rad/s)in elocity (angular v
)rad/sin accelation(angular constant
22
2
2
ifif
iif
if
tt
t
rrv
a
ra
rv
c
t
22
Physics 201: Lecture 23, Pg 7
Chapter 10
Physics 201: Lecture 23, Pg 8
Chapter 10
Physics 201: Lecture 23, Pg 9
Chapter 11
dtLd
Physics 201: Lecture 23, Pg 10
Chapter 12
Physics 201: Lecture 23, Pg 11
Approach to Statics:
In general, we can use the two equations
to solve any statics problems.
When choosing axes about which to calculate torque, choose one that makes the problem easy....
However if there is acceleration, then restrict rotation axis to center of mass (as well as for translation).
0F
0
Physics 201: Lecture 23, Pg 12
Young’s modulus: measures the resistance of a solid to a change in its length.
Changes in length: Young’s modulus
L0 L
Felasticity in length
0//
strain tensilestress tensileY
LLAF
A
Physics 201: Lecture 23, Pg 13
Bulk modulus: measures the resistance of solids or liquids to changes in their volume.
Changes in volume: Bulk Modulus
V0
Vi + V
F
Volume elasticity
iΔV/VF/AB
PressureF/AP
Physics 201: Lecture 23, Pg 14
Changes in shape: Shear Modulus
hx
AF
/
/
strainshear
stressshear S
Applying a force perpendicular to a surface
Physics 201: Lecture 23, Pg 15
Comparison Kinematics
Angular Linear
constant
0 t
200 2
1tt
constanta
at 0vv
221
00 v attxx
22
02
)( 021
AVE
ax2vv 2
02
)vv(v 021
AVE
Physics 201: Lecture 23, Pg 16
Comparison: Dynamics
Angular Linear
K 1
2I 2
K 1
2mv2
I = i mi ri2 m
F = a mr x F = I
L = r x = I p = mv
EXT dLdt
FEXT dp
dt
W = W = F •x
K = WNET K = WNET
Physics 201: Lecture 23, Pg 17
Momentum and collisions
Remember vector components
A 5 kg cart rolling without friction to the right at 10 m/s collides and sticks to a 5 kg motionless block on a 30° frictionless incline.
How far along the incline do the joined blocks slide?
Physics 201: Lecture 23, Pg 18
Momentum and collisions Remember vector components
A 5 kg cart rolling without friction to the right at 10 m/s collides and sticks to a 5 kg motionless block on a 30° frictionless incline.
Momentum parallel to incline is conserved Normal force (by ground on cart) is to the incline
mvi cos + m 0 = 2m vf vf = vi cos / 2 = 4.4 m/s
Now use work-energy
2mgh + 0 = ½ 2m v2f d = h / sin = v2
f / (2g sin )
mvi
mvi cos
Physics 201: Lecture 23, Pg 20
Example problem: Going in circles B A 2.0 kg disk tied to a long string undergoes circular motion on
a frictionless horizontal table top. The string passes through a hole and then hangs vertically. The disk starts out at 5.0 rev/sec 0.50 m away from hole. If you pull slowly down on the string so that the final radius is 0.25 m, what is the final angular velocity?
No external torque so angular momentum is conserved. Ib b = Ia a
I = mR2
(0.50)2 b = (0.25)2 a
4 b = a = 20 rev/sec ri
i
Physics 201: Lecture 23, Pg 21
Spinning ball on incline
A solid disk of mass m and radius R is spinning with angular velocity . It is positioned so that it can either move directly up or down an incline of angle (but it is not rolling motion). The coefficient of kinetic friction is . At what angle will the disk’s position on the incline not change?
Physics 201: Lecture 23, Pg 22
Spinning ball on incline A solid disk of mass m and radius R is spinning with
angular velocity . It is positioned so that it can either move directly up or down an incline of angle (but it is not rolling motion). The coefficient of kinetic friction is .
At what angle will the disk’s position on the incline not change?
0 F
mg
N
fsin0 mgfFx
cos0 mgNFy
cosmgNf
tan
Physics 201: Lecture 23, Pg 23
Spinning ball on incline A solid disk of mass m and radius r is spinning with angular
velocity . It is positioned so that it can either move directly up or down an incline of angle (but it is not rolling motion). The coefficient of kinetic friction is . While spinning the disk’s position will not change.
How long will it be before it starts to roll? This will occur only when =0.
CMdisk about 0 z
mg
N
f90sinsin|||| rffrz
cosmgrrfI rg /cos2
rgttf /cos20
Physics 201: Lecture 23, Pg 24
Example Problem A 3.0 kg mass is attached to a light, rigid rod 1.5 m long. The
rod is vertical and is anchored to the ground through a frictionless pivot. It sits perfectly balanced at an unstable equilibrium. A 500 gm bullet is shot horizontally at 100 m/s through the mass. The force versus time plot is shown.
How fast is the bullet going when
it leaves? What is the tension in the rod just
after the bullet exits?
Method: Impulse
I = area under curve
dttFI xx )(
dttI x ])15.0(10225[ 24
Physics 201: Lecture 23, Pg 25
Example Problem A 3.0 kg mass is attached to a light, rigid rod 1.5 m long. The
rod is vertical and is anchored to the ground through a frictionless pivot. It sits perfectly balanced at an unstable equilibrium. A 500 gm bullet is shot horizontally at 100 m/s through the mass. The force versus time plot is shown.
How fast is the bullet going when
it leaves?
dttI x ])15.0(10225[ 24
3.00.0
34 |3/)15.0(10225 ttI x
Ns 45xI
Ns)451000.5(
Ns 45
if PP
m/s 10 Ns 5 ff vP
Physics 201: Lecture 23, Pg 26
Example Problem A 3.0 kg mass is attached to a light, rigid rod 1.5 m long. The
rod is vertical and is anchored to the ground through a frictionless pivot. It sits perfectly balanced at an unstable equilibrium. A 500 gm bullet is shot horizontally at 100 m/s through the mass. The force versus time plot is shown.
What is the tension in the rod just
after the bullet exits?
m/s 15345
Ns 45
/ v
P
f
f
Tmgr
vmmaF cy
2
]N 305.1/)15(3[ 2 T
N 420N 30)(450 T
Physics 201: Lecture 23, Pg 27
Statics: Example
A sign of mass M is hung 1 m from the end of a 4 m long beam (mass m) as shown in the diagram. The beam is hinged at the wall. What is the tension in the wire in terms of m, M, g and any other given quantity?
wire
1 m
SIGN
Physics 201: Lecture 23, Pg 28
Statics: Example
mgMg
T
Fx
30°
Process: Make a FBD and note known / unknown forces.
Chose axis of rotation at support because Fx & Fy are not known
Fy
2 m
3 m
F = 0 0 = Fx – T cos 30°
0 = Fy + T sin 30° - mg - Mg
z-dir: z = 0 = -mg 2r – Mg 3r + T sin 30° 4r (r = 1m)
The torque equation get us where we need to go, T
T = (2m + 3M) g / 2
X
Physics 201: Lecture 23, Pg 29
Center of Mass Example: Astronauts & Rope
Two astronauts are initially at rest in outer space and 20 meters apart. The one on the right has 1.5 times the mass of the other (as shown). The 1.5 m astronaut wants to get back to the ship but his jet pack is broken. There happens to be a rope connected between the two. The heavier astronaut starts pulling in the rope.
(1) Does he/she get back to the ship ?
(2) Does he/she meet the other astronaut ?
M = 1.5mm
Physics 201: Lecture 23, Pg 30
Example: Astronauts & Rope(1) There is no external force so if the larger astronaut
pulls on the rope he will create an impulse that accelerates him/her to the left and the small astronaut to the right. The larger one’s velocity will be less than the smaller one’s so he/she doesn’t let go of the rope they will either collide (elastically or inelastically) and thus never make it.
M = 1.5mm
Physics 201: Lecture 23, Pg 31
Example: Astronauts & Rope
(2) However if the larger astronaut lets go of the rope he will get to the ship. (Too bad for the smaller astronaut!)
In all cases the center of mass will remain fixed because they are initially at rest and there is no external force.
To find the position where they meet all we need do is find the Center of Mass
M = 1.5mm
Physics 201: Lecture 23, Pg 32
Forces and rigid body rotation
To change the angular velocity of a rotating object, a force must be applied
How effective an applied force is at changing the rotation depends on several factorsThe magnitude of the forceWhere, relative to the axis of rotation the force is appliedThe direction of the force
F F
Which applied force will cause the wheel to spin the fastest?
A B C
Physics 201: Lecture 23, Pg 33
Leverage
The same concept applies to leveragethe lever undergoes rigid body rotation about a pivot
point:
F
F
F
Which applied force provides the greatest lift ?
AC
B
Physics 201: Lecture 23, Pg 34
More on torques
You need to change the tire on you car. You use a tire wrench which allows you to apply a pair of forces.
(A) What is the torque produced by a tire wrench of length L, given an applied couple of magnitude F, acting on a lug nut (point F) as shown in the figure?
(B) Assume the lug nut is stuck What is the torque acting on the wheel, if the lug nut is a distance r from the center?
Image courtesy John Wiley & Sons, Inc.
Physics 201: Lecture 23, Pg 35
Wheel wrench
F = (L/2) F + (L/2) F = LF
F = r F sin + r F sin ( )= = LF
3. F = L F + 0 F
Notice the torque is the same everywhere.
Physics 201: Lecture 23, Pg 36
For Thursday
Chapter 13 (Newton’s Law of Gravitation)
Physics 201: Lecture 23, Pg 37
Momentum & Impulse
A. 0
B. ½ v
C. 2 v
D. 4 v
A rubber ball collides head on (i.e., velocities are opposite) with a clay ball of the same mass. The balls have the same speed, v, before the collision, and stick together after the collision. What is their speed immediately after the collision?
Physics 201: Lecture 23, Pg 38
Momentum & Impulse
A rubber ball collides head on with a clay ball of the same mass. The balls have the same speed, v, before the collision, and stick together after the collision. What is their speed after the collision?
(a) 0
(b) ½ v
(c) 2 v
(d) 4 v
Physics 201: Lecture 23, Pg 39
Momentum, Work and Energy
A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a frictionless surface so that the spring is compressed 0.20 m. When the block is released, it slides across the surface and collides with the 0.60 kg bob of a pendulum. The bob is made of clay and the block sticks to it. The length of the pendulum is 0.80 m. (See the diagram.)
To what maximum height above the surface will the ball/block assembly rise after the collision? (g=9.8 m/s2)
A. 2.2 cmB. 4.4 cmC. 11. cmD. 22 cmE. 44 cmF. 55 cm
Physics 201: Lecture 23, Pg 40
Momentum, Work and Energy
A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a frictionless surface so that the spring is compressed 0.20 m. When the block is released, it slides across the surface and collides with the 0.60 kg bob of a pendulum. The bob is made of clay and the block sticks to it. The length of the pendulum is .80 m. (See the diagram.)
To what maximum height above the surface will the ball/block assembly rise after the collision?
A. 2.2 cmB. 4.4 cmC. 11. cmD. 22 cmE. 44 cmF. 55 cm
Physics 201: Lecture 23, Pg 41
Momentum, Work and Energy ( Now with friction)
A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a surface.
If mstatic = 0.54, how far can the spring be compressed and the block remain stationary (i.e., maximum static friction)?
F = 0 = k u - f = k u - N
u = mg/k = 0.54 (0.40x10 N) / 270 N/m= 0.0080 m
Physics 201: Lecture 23, Pg 42
Momentum, Work and Energy ( Now with friction)
A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a surface. The spring is compressed 0.20 m
If mkinetic = 0.50 and the block is 9.8 m away from the unstretched spring, how high with the clay/block pair rise?
Emech (at collision) = Uspring + Wfriction = ½ k u2 - mg d 1/2 m v2 = 135(0.04)-0.50(0.40x10.)10.=(540-20) J=520 J v2 = 1040/0.40 m2/s2 Now the collision (cons. of momentum) and the swing.
Physics 201: Lecture 23, Pg 43
Momentum and Impulse
Henri Lamothe holds the world record for the highest shallow dive. He belly-flopped from a platform 12.0 m high into a tank of water just 30.0 cm deep! Assuming that he had a mass of 50.0 kg and that he stopped just as he reached the bottom of the tank, what is the magnitude of the impulse imparted to him while in the tank of water (in units of kg m/s)?
(a) 121
(b) 286
(c) 490
(d) 623
(e) 767
Physics 201: Lecture 23, Pg 44
Momentum and Impulse
Henri Lamothe holds the world record for the highest shallow dive. He belly-flopped from a platform 12.0 m high into a tank of water just 30.0 cm deep! Assuming that he had a mass of 50.0 kg and that he stopped just as he reached the bottom of the tank, what is the magnitude of the impulse imparted to him while in the tank of water (in units of kg m/s)?
(a) 121
(b) 286
(c) 490
(d) 623
(e) 767p = sqrt(2x9.8x12.3)x50
Physics 201: Lecture 23, Pg 45
Momentum & Impulse
Suppose that in the previous problem, the positively charged particle is a proton and the negatively charged particle, an electron. (The mass of a proton is approximately 1,840 times the mass of an electron.) Suppose that they are released from rest simultaneously. If, after a certain time, the change in momentum of the proton is p, what is the magnitude of the change in momentum of the electron?
(a) p / 1840
(b) p
(c) 1840 p
Physics 201: Lecture 23, Pg 46
Momentum & Impulse
Suppose that in the previous problem, the positively charged particle is a proton and the negatively charged particle, an electron. (The mass of a proton is approximately 1,840 times the mass of an electron.) Suppose that they are released from rest simultaneously. If, after a certain time, the change in momentum of the proton is p, what is the magnitude of the change in momentum of the electron?
(a) p / 1840
(b) p
(c) 1840 p
Physics 201: Lecture 23, Pg 47
Conservation of Momentum
A woman is skating to the right with a speed of 12.0 m/s when she is hit in the stomach by a giant snowball moving to the left. The mass of the snowball is 2.00 kg, its speed is 25.0 m/s and it sticks to the woman's stomach. If the mass of the woman is 60.0 kg, what is her speed after the collision?
(a) 10.8 m/s
(b) 11.2 m/s
(c) 12.4 m/s
(d) 12.8 m/s
Physics 201: Lecture 23, Pg 48
Conservation of Momentum
A woman is skating to the right with a speed of 12.0 m/s when she is hit in the stomach by a giant snowball moving to the left. The mass of the snowball is 2.00 kg, its speed is 25.0 m/s and it sticks to the woman's stomach. If the mass of the woman is 60.0 kg, what is her speed after the collision?
(a) 10.8 m/s
(b) 11.2 m/s
(c) 12.4 m/s
(d) 12.8 m/s
Physics 201: Lecture 23, Pg 49
Conservation of Momentum
Sean is carrying 24 bottles of beer when he slips in a large frictionless puddle. He slides forwards with a speed of 2.50 m/s towards a very steep cliff. The only way for Sean to stop before he reaches the edge of the cliff is to throw the bottles forward at 20.0 m/s (relative to the ground). If the mass of each bottle is 500 g, and Sean's mass is 72 kg, what is the minimum number of bottles that he needs to throw?
(a) 18 (b) 20 (c) 21 (d) 24 (e) more than 24
Physics 201: Lecture 23, Pg 50
Momentum and Impulse
A stunt man jumps from the roof of a tall building, but no injury occurs because the person lands on a large, air-filled bag. Which one of the following statements best describes why no injury occurs?
(a) The bag provides the necessary force to stop the person.
(b) The bag reduces the impulse to the person.
(c) The bag reduces the change in momentum.
(d) The bag decreases the amount of time during which the momentum is changing and reduces the average force on the person.
(e) The bag increases the amount of time during which the momentum is changing and reduces the average force on the person.
Physics 201: Lecture 23, Pg 51
Momentum and Impulse
A stunt man jumps from the roof of a tall building, but no injury occurs because the person lands on a large, air-filled bag. Which one of the following statements best describes why no injury occurs?
(a) The bag provides the necessary force to stop the person.
(b) The bag reduces the impulse to the person.
(c) The bag reduces the change in momentum.
(d) The bag decreases the amount of time during which the momentum is changing and reduces the average force on the person.
(e) The bag increases the amount of time during which the momentum is changing and reduces the average force on the person.
Physics 201: Lecture 23, Pg 52
Momentum and Impulse
Two blocks of mass m1 = M and m2 = 2M are both sliding towards you on a frictionless surface. The linear momentum of block 1 is half the linear momentum of block 2. You apply the same constant force to both objects in order to bring them to rest. What is the ratio of the two stopping distances d2/d1?
(a) 1/ 2 (b) 1/ 2½
(c) 1 (d) 2½ (e) 2 (f) Cannot be determined without knowing the masses of
the objects and their velocities.
Physics 201: Lecture 23, Pg 53
Momentum and Impulse
Two blocks of mass m1 = M and m2 = 2M are both sliding towards you on a frictionless surface. The linear momentum of block 1 is half the linear momentum of block 2. You apply the same constant force to both objects in order to bring them to rest. What is the ratio of the two stopping distances d2/d1?
(a) 1/ 2 (b) 1/ 2½
(c) 1 (d) 2½ (e) 2 (f) Cannot be determined without knowing the masses of
the objects and their velocities.
Physics 201: Lecture 23, Pg 54
Momentum and Impulse In a table-top shuffleboard game, a heavy moving puck collides with a
lighter stationary puck as shown. The incident puck is deflected through an angle of 20° and both pucks are eventually brought to rest by friction with the table. The impulse approximation is valid (i.e.,the time of the collision is short relative to the time of motion so that momentum is conserved).
Which of the following statements is correct?A. The collision must be inelastic because the pucks have different
masses.B. The collision must be inelastic because there is friction between the
pucks and the surface.C. The collision must be elastic because the pucks bounce off each other.D. The collision must be elastic because, in the impulse approximation,momentum is conserved.E. There is not enough information given to decide whether the collision iselastic or inelastic.
Physics 201: Lecture 23, Pg 55
Momentum and Impulse In a table-top shuffleboard game, a heavy moving puck collides with a
lighter stationary puck as shown. The incident puck is deflected through an angle of 20° and both pucks are eventually brought to rest by friction with the table. The impulse approximation is valid (i.e.,the time of the collision is short relative to the time of motion so that momentum is conserved).
Which of the following statements is correct?A. The collision must be inelastic because the pucks have different
masses.B. The collision must be inelastic because there is friction between the
pucks and the surface.C. The collision must be elastic because the pucks bounce off each other.D. The collision must be elastic because, in the impulse approximation,momentum is conserved.E. There is not enough information given to decide whether the collision iselastic or inelastic.
Physics 201: Lecture 23, Pg 56
Bill (mass M) is climbing a ladder (length L, mass m) that leans against a smooth wall (no friction between wall and ladder). A frictional force F between the ladder and the floor keeps it from slipping. The angle between the ladder and the wall is .
What is the magnitude of F as a function of Bill’s distance up the ladder?
Bill
Lm
F
Exercise: Ladder against smooth wall
Physics 201: Lecture 23, Pg 57
Ladder against smooth wall...
Consider all of the forces acting. In addition to gravity and friction, there will be normal forces Nf and Nw by the floor and wall respectively on the ladder.
First sketch the FBD
Mgd
L/2
F
mg
Nw
Nf
y
x
Again use the fact that FNET = 0 in both x and y directions:
x: Nw = F
y: Nf = Mg + mg
Physics 201: Lecture 23, Pg 58
Ladder against smooth wall...
Since we are not interested in Nw, calculate torques about an axis through the top end of the ladder, in the z direction.
Mg
L/2
m
F
mg
Nw
Nf
y
x Substituting: Nf = Mg + mg and
solve for F :
torque axis
d
cos
0 sin sin sin)(sin2
fNLFLMgdLmgL
Mm
MgF2L
d tan
Physics 201: Lecture 23, Pg 59
Example: Ladder against smooth wall
For a given coefficient of static friction s,the maximum force of friction F that can beprovided is sNf = s g(M + m).
The ladder will slip if F exceedsthis value.
m
F
dCautionary note:Cautionary note:
(1) Brace the bottom of ladders!
(2) Don’t make too big!
We have just calculated that
Mm
MgF2L
d tan